Chebyshev's inequality

**roshanhero** · July 15th 2009, 06:56 AM

If X is a non-negative random variable for which E(X) exists,show that for every t>0.
$P[X\geq t] \leq\frac {E(X)}{t}$

**CaptainBlack** · July 15th 2009, 09:45 AM

Originally Posted by roshanhero

If X is a non-negative random variable for which E(X) exists,show that for every t>0.
$P[X\geq t] \leq\frac {E(X)}{t}$

$\frac{E(X)}{t}=\int_0^{\infty} \frac{x}{t}p(x)\; dx = \int_0^t\frac{x}{t}p(x)\; dx + \int_t^{\infty}\frac{x}{t}p(x)\; dx$

......... $\ge \int_t^{\infty}\frac{x}{t}p(x)\; dx$

and as $\frac{x}{t}\ge 1$ for $x\ge t$ hence:

$\frac{E(X)}{t}\ge \int_t^{\infty}\frac{x}{t}p(x)\; dx\ge \int_t^{\infty}p(x)\; dx$

CB

**roshanhero** · July 15th 2009, 08:27 PM

How can we prove chebyshev inequality from this relation?

**matheagle** · July 15th 2009, 08:28 PM

And you didn't notice that under Mr Fantasy's link...
http://people.csail.mit.edu/ronitt/COURSE/S07/lec25.pdf
under your last post...
http://www.mathhelpforum.com/math-he...nequality.html
it's Theorem ONE, on page ONE.

**CaptainBlack** · July 15th 2009, 08:38 PM

Originally Posted by roshanhero

How can we prove chebyshev inequality from this relation?

Put $X=(Y-\mu)^2$

CB

**roshanhero** · July 15th 2009, 08:45 PM

Thanks,but,what should i use in t.

**CaptainBlack** · July 15th 2009, 09:22 PM

Originally Posted by roshanhero

Thanks,but,what should i use in t.

try it, see what you get then refer th Chebyshev's inequality and see what you have to do with t to get the required form.

CB

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